Answer to Question #294819 in Statistics and Probability for Kristel Tabil

Question

construct the probability distribution of the random variable for the given situation. Two balanced diced are rolled. Let S be the random variable denoting the sum of the number of dots that will appear

Answer 1

Find the probability that the sum is 2:

$P(2) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{{36}}$ ((1,1) - the first and second dice rolled one point each)

Find the probability that the sum is 3:

$P(3) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = \frac{2}{{36}} = \frac{1}{{18}}$ ((1,2) or (2,1) - 1 point was dropped on the first dice, and 2 points on the second, or vice versa)

Find the probability that the sum is 4:

$P(4) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} = \frac{3}{{36}} = \frac{1}{{12}}$ ((1,3) or (3,1) or (2,2) - the first dice dropped 1 point, and the second 3 points, or vice versa, or both dice dropped 2 points)

Find the probability that the sum is 5:

$P(5) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = \frac{4}{{36}} = \frac{1}{{9}}$ ((1,4) or (4,1) or (2,3) or (3,2))

Find the probability that the sum is 6:

$P(6) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} = \frac{5}{{36}}$ ((1,5), (5,1), (2,4), (4,2), (3,3))

Find the probability that the sum is 7:

$P(7) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = \frac{6}{{36}} = \frac{1}{6}$ ((1,6) or (6,1) or (2,5) or (5,2) or (3,4) or (4,3))

Find the probability that the sum is 8:

$P(8) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} = \frac{5}{{36}}$ ((2,6) or (6,2) or (5,3) or (3,5) or (4,4))

Find the probability that the sum is 9:

$P(9) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = \frac{4}{{36}} = \frac{1}{9}$ ((3,6) or (6,3) or (5,4) or (4,5))

Find the probability that the sum is 10:

$P(10) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot \frac{1}{6} = \frac{3}{{36}} = \frac{1}{{12}}$ ((4,6) or (6,4) or (5,5))

Find the probability that the sum is 11:

$P(11) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = \frac{2}{{36}} = \frac{1}{{18}}$ ((5,6) or (6,5) )

Find the probability that the sum is 12:

$P(12) = \frac{1}{6} \cdot \frac{1}{6} \ = \frac{1}{{36}}$ (6,6 )

We have the probability distribution:

$\begin{matrix} X&2&3&4&5&6&7&8&9&{10}&{11}&{12}\\ p&{\frac{1}{{36}}}&{\frac{1}{{18}}}&{\frac{1}{{12}}}&{\frac{1}{9}}&{\frac{5}{{36}}}&{\frac{1}{6}}&{\frac{5}{{36}}}&{\frac{1}{9}}&{\frac{1}{{12}}}&{\frac{1}{{18}}}&{\frac{1}{{36}}} \end{matrix}$